Question

lesson 28 - exponential and logarithmic equations with applications
score: 50/100 answered: 10/20
question 10
score on last try: 0 of 4 pts. see details for more.
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use table and graph to find the solution accurate to one decimal place to the equation below
$2^x = 120$
6.7
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Explanation:

Step1: Recall logarithm property

To solve \(2^x = 120\), take the logarithm of both sides. Using the natural logarithm (ln), we get \(\ln(2^x)=\ln(120)\). By the power rule of logarithms, \(x\ln(2)=\ln(120)\).

Step2: Solve for x

Rearrange the equation to solve for \(x\): \(x = \frac{\ln(120)}{\ln(2)}\). Calculate \(\ln(120)\approx4.7875\) and \(\ln(2)\approx0.6931\). Then \(x\approx\frac{4.7875}{0.6931}\approx6.9\). (We can also check with a table or graph: when \(x = 6\), \(2^6 = 64\); \(x = 7\), \(2^7 = 128\). Since 120 is between 64 and 128, and closer to 128, we can refine. \(2^{6.9}=2^{7 - 0.1}=2^7\times2^{-0.1}\approx128\times0.933\approx119.4\), which is close to 120.)

Answer:

\(6.9\)